Problem: In triangle $XYZ$, we have $\angle Z = 90^\circ$, $XY = 10$, and $YZ = \sqrt{51}$. What is $\tan X$?
Answer: [asy]

pair X,Y,Z;

Z = (0,0);

Y = (sqrt(51),0);

X = (0,7);

draw(X--Y--Z--X);

draw(rightanglemark(Y,Z,X,15));

label("$X$",X,NE);

label("$Y$",Y,SE);

label("$Z$",Z,SW);

label("$10$",(X+Y)/2,NE);

label("$\sqrt{51}$",(Z+Y)/2,S);

[/asy]

Because this is a right triangle, $\tan X = \frac{YZ}{XZ}$.

Using the Pythagorean Theorem, we find $XZ = \sqrt{XY^2 - YZ^2} = \sqrt{100-51} = 7$.

So $\tan X = \boxed{\frac{\sqrt{51}}{7}}$.